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Full text of "Treatise On Analysis Vol-Ii"


4.    With the notation of Problems 2 and 3, show that if the moment problem for the
sequence (cn) has  two  distinct  solutions, then  both the series ]£ |Pn(A)|2 and

Z IQ*(A)|2 converge for J\ ^ 0 (use formula (4) of Problem 2).

Deduce that the moment problem has only one solution in each of the following
cases :

(a)   £ — = + oo       (Use formula (4) of Problem 3);

n   On

/u\    V>   r*« + l

(b)   2-7-T - =+°°

n   bnt>n + l

(use the relation

Pn(A)Qn+2(A) - Pn+2(X)Qn(X) = A,"VZn*1);

t>nOn + i

(c)    there exists a finite number r such that

£n-i + an + bn<r
for all n. (Note that equation (1) of Problem 3, with A = r, takes the form

bn(yn + i - y*) ~ bn- t(yn - yn-±) = (r - bn. l ~ an - bn)yn ,
and deduce that the sequence (Qn(r)) is an increasing sequence of numbers >0.)

5.    (a)   If (un) is a convergent series of real numbers >0, prove Carlemarfs inequality

by writing MiW2 ••*«„= (u^aiU2a2 ••• wn an)/(n + 1)" for suitably chosen ^n, to be
determined, and using the inequality of the means (Section 13.8, Problem 14).
(b)   With the notation of Problem 2, show that if

then the corresponding moment problem has only one solution (" Carleman's crite-
rion ")• (Observe that

n^ J(Pn«)2

"Pn« dv(t\

and deduce that b0bi • • • 6n_i g (c2n)1/2; then use (a).)

6. Let jET be a simple unbounded self-adjoint operator (Problem 1). Show that every
closed subspace of dom(jfiT) which is stable under H is of the form dom(B') n E(A),
where E(A) is the closed subspace of E which is the image of E under the orthogonal
projector <pA(H), .and A is a universally measurable subset of E. (By using 3(d) and
3(e), remark that we may assume that H is of the form A/v, where the bounded
measure v on R is such that polynomials are dense in

7. By using the existence of closed hermitian operators with defects (1, 0) or (0, 1)
(15.13.9), give examples of closed hermitian operators with defect (m, n) where m, n
are arbitrary integers ^0, or +°o.two forms